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3 years ago

Is muddy water a substance?

Physics

2 answers:

sergejj [24]3 years ago

7 0

Answer: No, I wouldn't imagine so.

Explanation:

Kazeer [188]3 years ago

7 0

Muddy water is not a substance

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Which of the circuit offers greater resistance to the flow of current 1A or 2A ?

Dafna1 [17]

1 A offers greater resistance

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3 years ago

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A capacitor is connected to an ac generator that has a frequency of 2.3 kHz and produces a rms voltage of 1.5 V. The rms current

miv72 [106K]

Answer:

Explanation:

Given that,

AC frequency of 2.3KHz

f=2.3×10³Hz

Vrms produce is

Vrms=1.5V

Current rms

Irms= 31mA

The capacitor is reconnected to a generator of frequency

f=4.8KHz =4800Hz

The current rms becomes

Irms= 85mA

Vrms=?

Solution

First genrator

The capacitive reactance is given as

Xc=Vrms/Irms

Xc=1.5/31×10^-3

Xc=48.39 ohms

Now, to know the capacitance of the capacitor

Xc=1/2πfC

Then,

C=1/2πfXc

So,

C=1/2×π×2300×48.39

C=1.43×10^-6C

C=1.43μF

Note: the capacitance of the capacitor did not change,

Now for generator two.

The reactance are given as

Xc=1/2πfC

Xc=1/2×π×4800×1.43×10^-6

Xc=23.19ohms

Then,

Vrms2=Irms2 ×Xc

Vrms2=85×10^-3×23.19ohms

Vrms2=1.97V

Vrms2=1.97Volts

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3 years ago

The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength $\lambda =1.4cm =0.014m$

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by $D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m$

So distance D will be 49.34 m

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3 years ago

What atmospheric conditions can cause sound wave refractions similar to those observed in the ocean?

Romashka-Z-Leto [24]

It has to do with we're the water is coming from and we're it's at like city or town

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3 years ago

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

<u><em> $t = 1.05s$ </em></u>

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3 years ago

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